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Question
Let ϕ(x) = f(x) + f(2a − x) and f"(x) > 0 for all x ∈ [0, a]. Then, ϕ (x)
Options
increases on [0, a]
decreases on [0, a]
increases on [−a, 0]
decreases on [a, 2a]
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Solution
Given: ϕ(x) = f(x) + f(2a − x)
Differentiating above equation with respect to x we get,
ϕ'(x) = f'(x) − f(2a − x) .....(1)
Since, f''(x) > 0, f'(x) is an increasing function.
Now,
when \[x \in \left[ 0, a \right]\]
\[f'\left( x \right) \leq f\left( 2a - x \right) . . . . . \left( 2 \right)\]
Considering equation (1) and (2) we get,
ϕ'(x) ≤ 0
⇒ ϕ'(x) is decreasing in [0, a]
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